[EM] Simplest paper count to produce a winner in the smith set.

[Jameson Quinn]

You can eliminate all candidates that have less than half the top approval
score, immediately. Australia is a bad example because they require full
ranking; but without that requirement, you could expect that each
candidate's approval total will be something on the order of min(50%,
2xFirst choices). Using the numbers you gave, that would mean only 3
parties left after elimination.

Jameson

2011/11/29 Clinton Mead <clintonmead at gmail.com>

>
>
> On Wed, Nov 30, 2011 at 2:47 AM, Kristofer Munsterhjelm <
> km_elmet at lavabit.com> wrote:
>
>> Clinton Mead wrote:
>>
>>> What would be the simplest paper count that will produce a winner in the
>>> smith set?
>>
>>
>> Then while there is more than one candidate left, eliminate the pairwise
>> loser of the two remaining candidates with the least approval score.
>>
>>
> I was thinking the a similar thing, but then I looked at typical election
> first preference counts, which are often like the following like the
> following (at least in Australia).
>
> Major Party 1: 41%
> Major Party 2: 39%
> Minor Party 1: 10%
> Minor Party 2: 3%
> Minor Party 3: 2%
> Minor Party 4: 2%
> Minor Party 5: 2%
> Minor Party 6: 1%
>
> Producing this entails one pass through the ballot (100% of papers need to
> be examined for one preference). This is the base line for FPP.
>
> For IRV, we can eliminate the 6 minor candidates in bulk, because they sum
> to less than the vote of the second candidate. So then we reexamine the 6
> minor party candidates for preferences. This means we have to look at 20%
> more votes.
>
> Our total work of IRV: 120% of FPP.
>
> For the method you suggest (with the minor difference that I'll use first
> preference to initially rank ballots) we need 100% work to do the first
> count, then, with Minor Party 6 and Minor Party 5 on a total of 3%, we need
> to look at the other 97% of ballots to do a pairwise comparison. Then we
> eliminate the pairwise loser, then we have to look at 95% of the remaining
> ballots to compare Minor 5/6 with Minor 4, etc.
>
> We get a result which might involve 400-500% more looks at ballot papers
> than FPP.
>
> The method I suggest is slightly different in that it starts from the top.
> We compare the top two candidates pairwise, this involves looking at 100%
> of the votes to do the first preference count, and an additional 20% to do
> the two candidate count. We're now at 120% of the work of FPP (one could
> say you could do the first preference and two candidate in the same pass,
> but this doesn't significantly lower the work, the time consuming work is
> looking for preferences, not flicking through ballot papers).
>
> Now, lets say Major Party 2 beats Major Party 1 pairwise. Then we
> distribute Major Party 1's preferences. This takes looking at 41% of the
> votes.
>
> Now we're at 161%.
>
> If Major Party 2 now has a majority, we have a winner. But if it doesn't,
> we pairwise compare Major Party 2 with Minor Party 1.
>
> This involves looking at all of the other minors vote, which is 10%. If
> Major Party 2 beats Minor Party 1, it probably has a majority, if it
> doesn't, it's very close, and will quickly get a majority as it defeats
> other minor candidates pairwise and distributes their preferences.
>
> If Minor Party 1 beats Major Party 2 however, then we need to distribute
> Major Party 2's preferences,  which involves looking at another 39-50% of
> the ballots.
>
> At this point, it is likely Minor Party 1 has a majority (particularly
> considering Major 1 and Major 2 are eliminated). If this is the case, we
> then just need to check Minor Party 1 v Major Party 1 pairwise. This
> involves looking at the 49% of votes which are not theirs. If Minor Party 1
> pairwise beats Major Party 1, it is the winner, if Major Party 1 beats
> Minor Party 1, we have a cycle (Major Party 1 < Major Party 2 < Minor Party
> 1 < Major Party 1) and we resolve this in favour of the first preference
> winner.
>
> The idea of this method is that it typically takes only 160% or so of the
> work of a FPP, and in the worse (typical) case perhaps 250% or at most 300%
> of the work of a FPP count. In particular, it's not particularly burdensome
> compared to IRV, which in the best case is around 120% and in the worse
> case around 200%.
>
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